Multiply the following complex numbers: $({1+i}) \cdot ({3-5i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({1+i}) \cdot ({3-5i}) = $ $ ({1} \cdot {3}) + ({1} \cdot {-5}i) + ({1}i \cdot {3}) + ({1}i \cdot {-5}i) $ Then simplify the terms: $ (3) + (-5i) + (3i) + (-5 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 3 + (-5 + 3)i - 5i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 3 + (-5 + 3)i - (-5) $ The result is simplified: $ (3 + 5) + (-2i) = 8-2i $